crowell - calculus (apr 2007)

1 Light and Matter Fullerton, California www.lightandmatter.com copyright 2005 Benjamin Crowell rev April 27, 2007 This book is licensed under the Creative Commons Attribution-ShareAlike license, version 1.0, http://creativecommons.org/licenses/by-sa/1.0/, except for those photographs and drawings of which I am not the author, as listed in the photo credits If you agree to the license, it grants you certain privileges that you would not otherwise have, such as the right to copy the book, or download the digital version free of charge from www.lightandmatter.com At your option, you may also copy this book under the GNU Free Documentation License version 1.2, http://www.gnu.org/licenses/fdl.txt, with no invariant sections, no front-cover texts, and no back-cover texts Rates of Change Probability, 57 1.1 Change in discrete steps Two sides of the same coin, 9.—Some guesses, 11 1.2 Continuous change 12 A derivative, 14.—Properties of the derivative, 15.— Higher-order polynomials, 16.—The second derivative, 16.—Maxima and minima, 18 Problems 21 Problems Techniques 4.1 4.2 4.3 4.4 Newton’s method Implicit differentiation Taylor series Methods of integration 65 66 67 72 Change of variable, 72.— Integration by parts, 74.— Partial fractions, 75 Problems To infinity — and beyond! 63 79 Complex number techniques 2.1 Infinitesimals 2.2 Safe use of infinitesimals 2.3 The product rule 2.4 The chain rule 2.5 Exponentials and logarithms 23 26 5.1 Review of complex 30 numbers 33 5.2 Euler’s formula 34 5.3 Partial fractions revisited Problems The exponential, 34.—The 81 84 86 87 logarithm, 35 2.6 Quotients 2.7 Differentiation computer 2.8 Continuity 2.9 Limits on 36 a Improper integrals 38 6.1 Integrating a function that 41 blows up 89 41 6.2 Limits of integration at infinity 90 L’Hˆpital’s o rule, 42.— Another perspective on inProblems 92 determinate forms, 44 Problems 46 Iterated integrals 7.1 Integrals inside integrals 93 7.2 Applications 95 3.1 Definite and indefinite 7.3 Polar coordinates 97 integrals 51 7.4 Spherical and cylindrical 3.2 The fundamental theorem of calculus 54 coordinates 98 3.3 Properties of the integral 55 Problems 100 3.4 Applications 56 Averages, 56.—Work, 57.— A Detours 103 Integration B Answers and solutions 111 C Photo Credits 129 D Reference 131 D.1 Review 131 Algebra, 131.—Geometry, area, and volume, 131.— Trigonometry with a right triangle, 131.—Trigonometry with any triangle, 131 D.2 Hyperbolic functions 131 D.3 Calculus 132 Rules for differentiation, 132.—Integral calculus, 132.—Table of integrals, 132 Preface thrust of the topic These details I’ve relegated to a chapter in the Calculus isn’t a hard subject back of the book, and the reader Algebra is hard I still remem- who has an interest in mathematber my encounter with algebra It ics as a career — or who enjoys a was my first taste of abstraction in nice heavy pot roast before moving mathematics, and it gave me quite on to dessert — will want to read a few black eyes and bloody noses those details when the main text suggests the possibility of a detour Geometry is hard For most people, geometry is the first time they have to proofs using formal, axiomatic reasoning I teach physics for a living Physics is hard There’s a reason that people believed Aristotle’s bogus version of physics for centuries: it’s because the real laws of physics are counterintuitive Calculus, on the other hand, is a very straightforward subject that rewards intuition, and can be easily visualized Silvanus Thompson, author of one of the most popular calculus texts ever written, opined that “considering how many fools can calculate, it is surprising that it should be thought either a difficult or a tedious task for any other fool to master the same tricks.” Since I don’t teach calculus, I can’t require anyone to read this book For that reason, I’ve written it so that you can go through it and get to the dessert course without having to eat too many Brussels sprouts and Lima beans along the way The development of any mathematical subject involves a large number of boring details that have little to with the main Rates of Change 1.1 Change in discrete steps Toward the end of the eighteenth century, a German elementary school teacher decided to keep his pupils busy by assigning them a long, boring arithmetic problem To oversimplify a little bit (which is what textbook authors always when they tell you about history), I’ll say that the assignment was to add up all the numbers from one to a hundred The children set to work on their slates, and the teacher lit his pipe, confident of a long break But almost immediately, a boy named Carl Friedrich Gauss brought up his answer: 5,050 b / A trick for finding the sum ing the area of the shaded region Roughly half the square is shaded in, so if we want only an approximate solution, we can simply calculate 72 /2 = 24.5 But, as suggested in figure b, it’s not much more work to get an exact result There are seven sawteeth sticking out out above the diagonal, with a total area of 7/2, so the total shaded area is (72 + 7)/2 = 28 In general, the sum of the first n numbers will be (n2 + n)/2, which explains Gauss’s result: (1002 + 100)/2 = 5, 050 Two sides of the same coin a / Adding the numbers from to Problems like this come up frequently Imagine that each household in a certain small town sends a total of one ton of garbage to the dump every year Over time, the garbage accumulates in the dump, taking up more and more space Figure a suggests one way of solving this type of problem The filled-in columns of the graph represent the numbers from to 7, and adding them up means find9 10 CHAPTER RATES OF CHANGE rate of change 13 n accumulated result 13n (n2 + n)/2 The rate of change of the function x can be notated as x Given the ˙ function x, we can always deter˙ mine the function x for any value of n by doing a running sum Likewise, if we know x, we can determine x by subtraction In the ˙ c / Carl Friedrich Gauss example where x = 13n, we can (1777-1855), a long time find x = x(n) − x(n − 1) = 13n − ˙ after graduating from ele13(n − 1) = 13 Or if we knew mentary school that the accumulated amount of Let’s label the years as n = 1, 2, garbage was given by (n + n)/2, 3, , and let the function1 x(n) we could calculate the town’s poprepresent the amount of garbage ulation like this: that has accumulated by the end of year n If the population is 2 constant, say 13 households, then n + n (n − 1) + (n − 1) − 2 garbage accumulates at a constant rate, and we have x(n) = 13n n2 + n − n2 + 2n − − n + = But maybe the town’s population =n is growing If the population starts out as household in year 1, and then grows to in year 2, and so on, then we have the same kind of problem that the young Gauss solved After 100 years, the accumulated amount of garbage will be 5,050 tons The pile of refuse grows more and more every year; the rate of change of x is not constant Tabulating the examples we’ve done so far, we have this: Recall that when x is a function, the notation x(n) means the output of the function when the input is n It doesn’t represent multiplication of a number x by a number n d / x is the slope of x ˙ The graphical interpretation of B 120 Answers and solutions where the first factor in brackets is the derivative of the function on the outside, and the second one is the derivative of the “inside stuff.” Simplifying a little, the answer is 200(2x + 3)99 page 47, problem 9: Applying the product rule, we get (x + 1)99 (x + 2)200 + (x + 1)100 (x + 2)199 (The chain rule was also required, but in a trivial way — for both of the factors, the derivative of the “inside stuff” was one.) page 47, problem 10: The derivative of e7x is e7x · 7, where the first factor is the derivative of the outside stuff (the derivative of a base-e exponential is just the same thing), and the second factor is the derivative of the inside stuff This would normally be written as 7e7x x The derivative of the second function is ee ex , with the second exponential factor coming from the chain rule page 47, problem 11: We need to put together three different ideas here: (1) When a function to be differentiated is multiplied by a constant, the constant just comes along for the ride (2) The derivative of the sine is the cosine (3) We need to use the chain rule The result is −ab cos(bx + c) page 47, problem 12: If we just wanted to fine the integral of sin x, the answer would be − cos x (or − cos x plus an arbitrary constant), since the derivative would be −(− sin x), which would take us back to the original function The obvious thing to guess for the integral of a sin(bx + c) would therefore be −a cos(bx + c), which almost works, but not quite The derivative of this function would be ab sin(bx + c), with the pesky factor of b coming from the chain rule Therefore what we really wanted was the function −(a/b) cos(bx + c) page 47, problem 13: To find a maximum, we take the derivative and set it equal to zero The whole factor of 2v /g in front is just one big constant, so it comes along for the ride To differentiate the factor of sin θ cos θ, we need to use the chain rule, plus the fact that the derivative of sin is cos, and the 121 derivative of cos is − sin 0= 2v (cos θ cos θ + sin θ(− sin θ)) g = cos2 θ − sin2 θ cos θ = ± sin θ We’re interested in angles between, and 90 degrees, for which both the sine and the cosine are positive, so cos θ = sin θ tan θ = θ = 45 ◦ To check that this is really a maximum, not a minimum or an inflection point, we could resort to the second derivative test, but we know the graph of R(θ) is zero at θ = and θ = 90 ◦ , and positive in between, so this must be a maximum page 47, problem 14: Taking the derivative and setting it equal to zero, we have (ex − e−x ) /2 = 0, so ex = e−x , which occurs only at x = The second derivative is (ex + e−x ) /2 (the same as the original function), which is positive for all x, so the function is everywhere concave up, and this is a minimum page 47, problem 15: g/A, so that d = A ln cosh ct = (a) As suggested, let c = A ln (ect + e−ct ) Applying the chain rule, the velocity is A cect − ce−ct cosh ct (b) The expression can be rewritten as Ac ct (c) For large t, the e−ct terms become negligible, so the velocity is Acect /ect = Ac (d) From the original expression, A must have units of distance, since the logarithm is unitless Also, since ct occurs inside a function, ct must be unitless, which means that c has units of inverse time The answers to parts b and c get their units from the factors of Ac, which have units of distance multiplied by inverse time, or velocity page 48, problem 16: Since I’ve advocated not memorizing the quotient rule, I’ll this one B 122 Answers and solutions from first principles, using the product rule d tan θ dθ d sin θ = dθ cos θ d −1 = sin θ (cos θ) dθ −1 = cos θ (cos θ) + (sin θ)(−1)(cos θ)−2 (− sin θ) = + tan2 θ (Using a trig identity, this can also be rewritten as sec2 θ.) page 48, problem 17: √ Reexpressing x as x1/3 , the derivative is (1/3)x−2/3 page 48, problem 18: (a) Using the chain rule, the derivative of (x2 + 1)1/2 is (1/2)(x2 + 1)−1/2 (2x) = x(x2 + 1)−1/2 (b) This is the same as a, except that the is replaced with an a2 , so the answer is x(x2 + a2 )−1/2 The idea would be that a has the same units as x (c) This can be rewritten as (a+x)−1/2 , giving a derivative of (−1/2)(a+ x)−3/2 (d) This is similar to c, but we pick up a factor of −2x from the chain rule, making the result ax(a − x2 )−3/2 page 48, problem 19: By the chain rule, the result is 2/(2t + 1) 123 page 48, problem 20: Using the product rule, we have d sin x + dx d sin x dx , but the derivative of a constant is zero, so the first term goes away, and we get cos x, which is what we would have had just from the usual method of treating multiplicative constants page 48, problem 21: N(Gamma(2)) N(Gamma(2.00001)) 1.0000042278 N( (1.0000042278-1)/(.00001) ) 0.4227799998 Probably only the first few digits of this are reliable page 48, problem 22: The area and volume are A = 2πr + 2πr2 and V = πr2 The strategy is to use the equation for A, which is a constant, to eliminate the variable , and then maximize V in terms of r = (A − 2πr2 )/2πr Substituting this expression for V = back into the equation for V , rA − πr3 B 124 Answers and solutions To maximize this with respect to r, we take the derivative and set it equal to zero A − 3πr2 A = 6πr2 0= = (6πr2 − 2πr2 )/2πr = 2r In other words, the length should be the same as the diameter page 48, problem 23: (a) We can break the expression down into three factors: the constant m/2 in front, the nonrelativistic velocity dependence v , and the relativistic correction factor (1 − v /c2 )−1/2 Rather than substituting in at for v, it’s a little less messy to calculate dK/dt = (dK/dv)(dv/dt) = adK/dv Using the product rule, we have dK = a · m 2v − dt +v · − = ma2 t + 1− 1− v2 1− 2c2 v2 c2 v2 c2 v2 c2 v2 c2 −1/2 −3/2 − 2v c2 −1/2 −3/2 (b) The expression ma2 t is the nonrelativistic (classical) result, and has the correct units of kinetic energy divided by time The factor in square brackets is the relativistic correction, which is unitless (c) As v gets closer and closer to c, the expression − v /c2 approaches zero, so both the terms in the relativistic correction blow up to positive infinity page 49, problem 24: We already know it works for positive x, so we only need to check it for negative x For negative values of x, the chain rule tells us that the derivative is 1/|x|, multiplied by −1, since d|x|/dx = −1 This gives −1/|x|, which is the same as 1/x, since x is assumed negative 125 page 49, problem 25: Let f = dxk /dx be the unknown function Then dx dx d = xk x−k+1 dx = f x−k+1 + xk (−k + 1)x−k 1= , where we can use the ordinary rule for derivatives of powers on x−k+1 , since −k + is positive Solving for f , we have the desired result Solutions for chapter page 63, problem 1: a := 0; b := 1; H := 1000; dt := (b-a)/H; sum := 0; t := a; While (t